# Question about Maclaurin series - calculus

1. Apr 27, 2015

### andydan1060

1. The problem statement, all variables and given/known data
Find the Maclaurin series of the function https://webwork.wustl.edu/webwork2_files/tmp/equations/87/63afd4b6f3566e2a90aa420dc5d1821.png [Broken]
c_3 =
c_4 =
c_5 =
c_6 =
c_7 =

2. Relevant equations

3. The attempt at a solution
(8x^2)[(9x) - (9x)^3/3! + (9x)^5/5! - (9x)^7/7! + ...]

I got c_3 = 72, which is correct, and c_5 = (8x^2)[(9x^3)/3!] = 5832/3! (coefficient)

Apparently that's wrong and it's supposed to be -116640/(5!).

What did I do wrong? Thanks!

Last edited by a moderator: May 7, 2017
2. Apr 27, 2015

### PeroK

Are you not just muddling up which coefficient is which? If you've got 1/3! and the answer has 1/5! then those are different coefficients.

3. Apr 27, 2015

### Delta²

Isnt c4 supposed to be the coefficient of x^4 which is zero in this series?

4. Apr 27, 2015

### andydan1060

It's supposed to be c_5, not c_4, sorry.

Last edited by a moderator: May 7, 2017
5. Apr 27, 2015

### PeroK

First, your coefficient should have a minus.

Second, why do you think your answer and the answer given are different? Just because they look different doesn't mean they are!

6. Apr 27, 2015

### Delta²

Well PeroK is right, doing the calculations it turns out that 5832/3!=116640/5!. However i wonder if your teacher wanted you to do this assignment by manually calculating up to the 7th derivative of 8x^2sin(9x) and evaluating the derivatives at 0.

7. Apr 27, 2015

### PeroK

I suspect that since we have the coefficient of $x^5$, whoever gave the answer preferred to have $5!$ in the denominator.